Permutations matches in basketball league

[master cherundo]

Homework Statement

Suppose that a basketball league has 32 teams, split into two conferences of 16 teams each. Each conference is split into three divisions. Suppose that the North Central Division has five teams. Each of the teams in the North Central Division plays four games agains each of the other teams in this division, three games against each of the 11 remaining teams in the conference, and two games against each of the 16 teams in the other conference. In how many different orders can the games of one of the teams in the North Central DIvision be scheduled?

Homework Equations

$$\frac{n!}{n_1n_2 \cdots n_k}$$

The Attempt at a Solution

Well, I think it would be $$\frac{81!}{4!^4 \times 11!^3 \times 16!^2}$$
But, it is a hard stuff to compute this if we do not use calculator. And, I believe it has implementation with how to distribute $$82$$ objects into $$n$$ boxes. Unfortunately, I do not know how to find $$n$$.

 

[Tedjn]

I don't think you are understanding what a permutation is exactly. The definition of a permutation is:

$$_nP_k = \frac{n!}{(n-k)!}$$

The reason for this is because, in the beginning, you have n objects from which to choose. For the second object, you have n-1 choices. Third object, you have n-2 choices. Kth object, you have n-k+1 choices. This multiplication is equivalent to the permutation definition above.

In your case, how many total games are there? Are the independent of each other--that is, does playing one game affect whether you can play any of the other games? Think about how many total choices you have for the first game, how many total choices you have for the second game, etc.

 

[Architeuthis Dux]

I would approach this problem by first understanding the concept of permutations and the formula for calculating them. In this case, we have 81 games to be scheduled for one team in the North Central Division. Using the formula for permutations, we have 81! possible ways to arrange these games. However, we must also consider the fact that the team will play four games against each of the other four teams in their division, three games against each of the 11 remaining teams in the conference, and two games against each of the 16 teams in the other conference. This means that we must divide by the number of ways these games can be arranged within each category.

For the games within the division, we have 5 teams and 4 games each, so we have 5!^4 ways to arrange these games. For the games within the conference, we have 11 remaining teams and 3 games each, so we have 11!^3 ways to arrange these games. And for the games in the other conference, we have 16 teams and 2 games each, so we have 16!^2 ways to arrange these games.

Therefore, the total number of possible schedules for one team in the North Central Division would be:

\frac{81!}{5!^4 \times 11!^3 \times 16!^2} = 4.331 \times 10^{74} possible schedules.

This is a very large number, but it is important to consider all the different factors and possibilities when scheduling games in a basketball league. As a scientist, it is important to carefully consider all the variables and use mathematical formulas to find the most accurate and efficient solution.